# Parametrization of one-sheet hyperboloid

I was given the following exercise: let $$S$$ be $$x^2 +y^2-z^2=1$$.

1. show that for every real number $$t$$ the line $$l_t$$ $$(x-z)\cos t=(1-y)\sin t,\quad (x+z)\sin t=(1+y)\cos t$$ is contained in $$S$$;
2. show that every point of $$S$$ is contained in one and only one of the above lines;
3. use this remark to parametrize $$S$$.

My approach was to observe that $$l_t = l_{t+k\pi}$$; then I defined $$t:=\arctan\left(\frac{1+y}{x+z}\right)$$ and showed that $$p=(x,y,z) \in S \iff p \in l_t\text{,}$$ obviously assuming $$t\neq \pm \frac{\pi}{2}, y\neq 1, x\neq -z$$. The 'only one' part was not a problem. Finally I wrote $$l_t$$ in parametric form obtaining a parametrization for $$S$$.

My question is: is there a way to solve the exercise so that the parametrization obtained for $$S$$ is unique? Unfortunately in my solution I have to consider an atlas of parametrizations, since only one parametrization covers $$S$$ minus a line (e.g. $$y=1, x=-z$$).

Thank you.

• "contained in one and only one of the above lines": this is a little contradictory as $t$ is unbounded and the lines are repeated infinitely many times.
– user65203
Jun 12, 2020 at 9:25

Your lines $$l_t$$ are presented in a strange way. Solving the two equations for $$x$$ and $$y$$ gives l_t:\qquad\eqalign{x&=\sin(2t)+\cos(2t)\> z,\cr y&=-\cos(2t)+\sin(2t)\> z\ .\cr} I'm taking the freedom to replace your parameter $$t$$ that "numbers" the lines $$l_t$$ by $$\phi:=2t-{\pi\over2}\quad(0\leq t<\pi),\qquad{\rm resp.}\qquad t={1\over2}\left(\phi+{\pi\over2}\right)\quad(0\leq\phi<2\pi)\ .$$ I'm then talking about the lines $$\ell_\phi: \quad z\mapsto\bigl(\cos\phi-\sin\phi \>z,\ \sin\phi+\cos\phi \>z,\ z\bigr)\qquad(-\infty with $$0\leq\phi<2\pi$$. When $$z=0$$ the line $$\ell_\phi$$ passes through the point $$(\cos\phi,\sin\phi),0)$$ of the unit circle in the $$(x,y)$$-plane. In fact the projection $$\ell'_\phi: \quad z\mapsto\bigl(\cos\phi-\sin\phi \>z,\ \sin\phi+\cos\phi \>z,\ 0\bigr)\qquad(-\infty of the $$\ell_\phi$$ into the $$(x,y)$$-plane is just the tangent to the unit circle, $$\phi$$ being the polar angle of the point of tangency. When I change the value of $$\phi$$ to a $$\phi'\ne\phi$$ (modulo $$2\pi$$) I obtain a different tangent, and it is then geometrically obvious that the lines $$\ell_\phi$$ and $$\ell_{\phi'}$$ don't intersect in space.
The new parametrization of $$S$$ now comes for free from $$(1)$$. We just write S:\quad (\phi,z)\mapsto\left\{\eqalign{x&=\cos\phi-\sin\phi \>z\cr y&=\sin\phi+\cos\phi \>z\cr z&=z\cr}\right.\qquad\qquad(\phi\in{\mathbb R}/(2\pi), \ -\infty
Taking $$y$$ as a free variable,
$$x-z=(1-y)\tan(t)\\x+z=(1+y)\cot(t)$$ give you a parameterization in terms of $$y$$ and $$t$$. If $$t$$ is contrained to $$[0,\pi)$$, the parameterization is unique from the result 2. and the fact that we intersect a straight line with a plane.
(The special cases such that $$\tan(t)$$ or $$\cot(t)$$ are undefined correspond to the two lines $$y=\pm1, x=\pm z$$.)